Đáp án:
Gọi $UCLN(8a+3b;5a+2b)=d$
$→\left\{\begin{matrix}
8a+3b\qquad\vdots\qquad d & \\
5a+2b \qquad\vdots\qquad d&
\end{matrix}\right.$
$\to \left\{\begin{matrix} 5(8a+3b)\qquad\vdots\qquad d & \\ 8(5a+2b) \qquad\vdots\qquad d& \end{matrix}\right.$
$\to \left\{\begin{matrix} 40a+15b\qquad\vdots\qquad d & \\ 40a+16b \qquad\vdots\qquad d& \end{matrix}\right.$
$\to b\qquad\vdots\qquad d \qquad(1)$
$\left\{\begin{matrix} 8a+3b\qquad\vdots\qquad d & \\ 5a+2b \qquad\vdots\qquad d& \end{matrix}\right.$
$\to\left\{\begin{matrix} 2(8a+3b)\qquad\vdots\qquad d & \\ 3(5a+2b) \qquad\vdots\qquad d& \end{matrix}\right.$
$\to\left\{\begin{matrix} 16a+6b\qquad\vdots\qquad d & \\ 15a+6b\qquad\vdots\qquad d& \end{matrix}\right.$
$\to a\qquad\vdots\qquad d \qquad (2)$
Từ $(1);(2)→d=1\to \dfrac{8a+3b}{5a+2b}$ tối giản ( vì $a; b$ nguyên tố cùng nhau )
